Tiêu đề Basic Mathematics K Scheme Notes by mypractically.pdf ✅

Basic Mathematics (22103) 5 | P a g e Unit 1 Algebra Course Outcome: Apply the concepts of algebra to solve engineering related problems. Unit outcome: a) Solve the given simple problem based on laws of logarithm. b) Calculate the area of the given triangle by determinant method. c) Solve given system of linear equations using matrix inversion method and by Cramer’s rule. d) Obtain the proper and improper partial fraction for the given simple rational function. Introduction: Algebra is a simple language, used to create mathematical models of real world situations and to handle problems. The algebraic need of engineering and technology is to solve simple engineering problems using algebra. Some of the main topics coming algebra includes Logarithms, Determinants, Matrix and Partial fractions. Logarithm Significance of Logarithms: Logarithm is one of the best tools to simplify engineering problems. Content of Logarithms: Definition: If y = ax , a > 0, a  1, a  R, then x is called logarithm of y to the base a and it is written as x = loga y. For example, 1) If 8 = 2 3 then 3 = log2 8 2) If 3 4 = 81 then log3 81 = 4 Note: i) ax = y is called Exponential form or Index form and x = loga y is called Logarithmic form of the same expression. ii) Logarithm of negative number and zero are not defined LAWS OF LOGARITHM: 1. loga (m n) = loga m + loga n 2. loga      m n = loga m  loga n
Basic Mathematics (22103) 6 | P a g e 3. loga (m) n = n loga m 4. logn m = loga m loga n Remark: 1. a 0 =1  loga 1 = 0 2. a 1 = a  loga a = 1 3. a log a y = y Solved Examples: Evaluate the following a) log 216 Solution: log 2 16 = log 2 2 4 = 4 log 2 2 = 4(1) = 4 b) log 5 125 Solution: log 5 125 = log 5 5 3 = 3 log 5 5 = 3(1) = 3 c) 25log5 8 Solution : 25log5 8 =[(5) 2 ] log5 8 = 52 log 5 8 = 5 log 5 8 2 = 5 log 5 64... (aloga y = y) = 64 Simplify the following a) log 2 14 log 2 7 Solution: log 2 14 log 2 7
Basic Mathematics (22103) 7 | P a g e        7 14 log 2  log2 2 = 1 b) (log 4) (log 81) 3  4 Solution: (log 4) (log 81) 3  4 = log 4 log 81 log 3 log 4  log 3 log 81  log 3 log 3 4  log 3 log 3  4 = 4 c)                     15 8 log 5 4 log 3 2 log Solution:                     15 8 log 5 4 log 3 2 log =               15 8 log 5 4 3 2 log =              15 8 log 15 8 log =        8 15 15 8 log = log1 = 0 Find x if i) log3 27 = x Solution: log3 27 = x 3 x = 27 3 x = 33  x = 3